Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
vector<int> nd[maxn][20], eg[maxn], c[maxn], s[maxn];
int d[maxn], pre[maxn][20];
int n;
void Init() {
  memset(pre, 0, sizeof(pre));
  memset(d, 0, sizeof(d));
}
void Dfs(int u, int pr) {
  pre[u][0] = pr;
  d[u] = d[pr] + 1;
  if (pr != 0 && s[pr].size())
    for (int i = 0; i < s[pr].size(); i++) nd[u][0].push_back(s[pr][i]);
  for (int i = 0; i < eg[u].size(); i++) {
    int v = eg[u][i];
    if (v != pr) Dfs(v, u);
  }
}
vector<int> MergeSort(vector<int>& v1, vector<int>& v2, int a = 10) {
  int l1 = v1.size(), l2 = v2.size();
  int l = min(a, l1 + l2);
  vector<int> ret;
  int p1 = 0, p2 = 0;
  while (ret.size() < l) {
    if (p1 < l1 && p2 < l2) {
      if (v1[p1] <= v2[p2])
        ret.push_back(v1[p1++]);
      else
        ret.push_back(v2[p2++]);
    } else {
      if (p1 < l1) ret.push_back(v1[p1]), p1++;
      if (p2 < l2) ret.push_back(v2[p2++]);
    }
  }
  return ret;
}
void Merge(int u, int c) {
  vector<int> ret = MergeSort(nd[u][c - 1], nd[pre[u][c - 1]][c - 1]);
  for (int i = 0; i < ret.size(); i++) nd[u][c].push_back(ret[i]);
}
void Lca_Init() {
  for (int i = 1; i < 20; ++i)
    for (int j = 1; j <= n; j++) {
      pre[j][i] = pre[pre[j][i - 1]][i - 1];
      if (pre[j][i]) Merge(j, i);
    }
}
void Print(vector<int>& ans, int k) {
  int t = min(k, (int)ans.size());
  printf("%d", t);
  for (int i = 0; i < t; i++) printf(" %d", ans[i]);
  printf("\n");
}
void Lca(int x, int y, int k) {
  if (x == y) {
    Print(s[x], k);
    return;
  }
  vector<int> ret = MergeSort(s[x], s[y], k);
  if (d[x] > d[y]) swap(x, y);
  if (d[y] > d[x]) {
    for (int i = 19; i >= 0; i--)
      if (d[pre[y][i]] > d[x]) {
        ret = MergeSort(ret, nd[y][i]);
        y = pre[y][i];
      }
  }
  if (pre[y][0] == x) {
    Print(ret, k);
    return;
  }
  if (d[y] > d[x]) ret = MergeSort(ret, nd[y][0]), y = pre[y][0];
  for (int i = 19; i >= 0; i--) {
    if (pre[x][i] != pre[y][i]) {
      ret = MergeSort(ret, nd[x][i]);
      ret = MergeSort(ret, nd[y][i]);
      x = pre[x][i], y = pre[y][i];
    }
  }
  ret = MergeSort(ret, nd[x][0]);
  Print(ret, k);
}
int main() {
  Init();
  int m, q;
  scanf("%d%d%d", &n, &m, &q);
  int x, y;
  for (int i = 1; i <= n - 1; i++) {
    scanf("%d%d", &x, &y);
    eg[x].push_back(y), eg[y].push_back(x);
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d", &x);
    c[x].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    if (c[i].size() > 0) sort(c[i].begin(), c[i].end());
    for (int j = 0; j < min(10, (int)c[i].size()); j++) s[i].push_back(c[i][j]);
  }
  Dfs(1, 0);
  Lca_Init();
  int k;
  while (q--) {
    scanf("%d%d%d", &x, &y, &k);
    Lca(x, y, k);
  }
  return 0;
}